A)compute the mole of AgNO3 first.
volume, when V=1.92 L
use:
amount of mole,
n = Volume * Molarity
= 0.18*1.92
= 0.3456 mole
In light of the balanced equation
AgNO3 = (2/2)*moles of AgCl produced
= (2/2)*0.3456
= 0.3456 mole
This is the quantity of AgCl moles.
AgCl molar mass,
MM equals 1*MM(Ag) + 1*MM (Cl)
= 1*107.9 + 1*35.45
= 143.35 g/mole
use: volume of AgCl,
M = Molar Mass * Molar Number
= 1.434*102 mole/mole x 0.3456 mole
= 49.54 g
Response: 49.5 g
B)Here: AgNO3 = M=0.18 M
V(AgNO3)=1.92 liters
V (MgCl2) = 3.83 L
A balanced response states:
1 mole of AgNO3 = 2 mole of MgCl2, and vice versa.
2*M(MgCl2)*V = 1*M(AgNO3)*V(AgNO3) (MgCl2)
1*0.18*1.92 = 2*M(MgCl2)*3.83
0.0451 M (MgCl2)
Response: 0.0451 M
To know more about magnesium chloride, click on the link below:
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